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I have two sequences x and y of lengths, say 5 and 10. I multiply them in time domain element by element. I get another sequence. Now this as per the convolution theorem should be equal to the inverse z tranform of the convolution of the two z transforms of the original sequences. Now, I know how to get the ZTs of the original sequences. How do I get the convolution of these ZTs?

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  • $\begingroup$ As usual, I assume that the sequences are infinite in length, the non-given values are zeros and so I pad with zeros to make the lengths equal and proceed further with calculations. $\endgroup$ – Seetha Rama Raju Sanapala Apr 30 '15 at 13:19
  • $\begingroup$ It's not clear where you're running into trouble; try giving more details. As you noted, you need to convolve the two sequence, then inverse transform the result. You didn't really specify what part you don't understand. $\endgroup$ – Jason R Apr 30 '15 at 13:20
  • $\begingroup$ Thanks for the response. I got stuck up how to find the convolution in z-domain once I find the ZTs. $\endgroup$ – Seetha Rama Raju Sanapala Apr 30 '15 at 13:22
  • $\begingroup$ Generally we find convolution in domain by finding the multiplication of ZTs. I am trying to do the reverse. Find (actually checking) the time multiplicaiton using convolution of ZTs. $\endgroup$ – Seetha Rama Raju Sanapala Apr 30 '15 at 13:24
  • $\begingroup$ Do you have a formula for the convolution of two $z$-transforms that is more detailed than $$(G\star H)(z) = G(z)\star H(z)?$$ that is, something comparable to $$(G\star H)(f) = G(f)\star H(f) = \int_{-\infty}^\infty G(f-\lambda)H(\lambda)\,\mathrm d\lambda$$ for the convolution of two Fourier transforms? and do you know how to use the formula? $\endgroup$ – Dilip Sarwate Apr 30 '15 at 15:14
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The term-by-term product of the sequences $(a_0, a_1, a_2)$ and $(b_0, b_1, b_2)$ is $(c_0, c_1, c_2) = (a_0b_0, a_1b_1, a_2b_2)$. The corresponding $z$-transforms are \begin{align} A(z) &= a_0 + a_1z^{-1} + a_2z^{-2}\\ B(z) &= b_0 + b_1z^{-1} + b_2z^{-2}\\ C(z) &= a_0b_0 + a_1b_1z^{-1} + a_2b_2z^{-2} \end{align} According to the formula in Wikipedia pointed out to you by MattL, $C(z)$ can be obtained from $A(z)$ and $B(z)$ via a contour integral in the complex plane. The integrand of the contour integral (which is with respect to the complex variable $v$) is \begin{align} v^{-1}A(v)B\left(\frac zv\right) &= v^{-1}\left(a_0 + a_1v^{-1} + a_2v^{-2}\right) \left(b_0 + b_1\left(\frac zv\right)^{-1} + b_2\left(\frac zv\right)^{-2}\right)\\ &= \frac {a_0b_0 + a_1b_1z^{-1} + a_2b_2z^{-2}}{v} + \sum_{m=0}^2\sum_{n=0, n\neq m}^2 \left(a_mb_nz^{-n}\right) v^{-m + n -1}\tag{1} \end{align} where in that last sum in $(1)$, the exponent of $v$ in each term is either $\geq 0$ or $\leq -2$. In other words, if we regard each of the $8$ terms in that sum as a separate function of $v$, then the function either has no poles inside the unit circle ($-m+n-1 \geq 0$) or has a pole of multiplicity greater than $1$ at the origin ($-m+n-1 \leq -2$). Now, write the contour integral of the right side of $(1)$ as the sum of $9$ different contour integrals and apply Cauchy's integral theorem to evaluate each of these $9$ contour integrals. The conclusion is that the integrals of all the terms in that sum in $(1)$ integrate to $0$ while the contour integral of $\displaystyle \frac {a_0b_0 + a_1b_1z^{-1} + a_2b_2z^{-2}}{v}$ (which has a single pole at the origin) gives the desired $C(z) = a_0b_0 + a_1b_1z^{-1} + a_2b_2z^{-2}$. Generalization of all this to sequences of arbitrary lengths is straightforward.

If you don't know enough complex-variable theory to figure out why it is that the contour integral of the first term in $(1)$ is just the numerator and why the integral of the other terms is $0$, then you will not be able to carry out the verification that you desire.

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  • $\begingroup$ It may be helpful to add that strictly speaking $D(v)$ is not a polynomial in $v$ because it also contains negative powers of $v$, and it is not independent of $z$, as the notation might suggest to the uninitiated. The conclusion of course remains true, because the only important thing is that $D(v)$ does not contain terms with $v^{-1}$. $\endgroup$ – Matt L. May 1 '15 at 10:59
  • $\begingroup$ @MattL. Thanks. It was late at night when I wrote my answer and forgot to include the poles at the origin of multiplicity greater than $1$. have edited my answer to point all this out explicitly. $\endgroup$ – Dilip Sarwate May 1 '15 at 13:10
  • $\begingroup$ OK, I think this is clearer now. I guess the sum is meant as a double sum over $n$ and $m$ where both indices vary between $0$ and $2$, with $m\neq n$. $\endgroup$ – Matt L. May 3 '15 at 9:28
  • $\begingroup$ @MattL. Thanks. Yes, that is a double sum. I have edited the answer to include this as well as more explanations. $\endgroup$ – Dilip Sarwate May 3 '15 at 13:50

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